Question

If V be the volume of a tetrahedron and \[V'\] be the volume of another tetrahedron formed by the centroids of faces of the previous tetrahedron and\[V = KV'\] , then K is equal to
a. $9$
b. $12$
c. $27$
d. $81$

Answer 1

Hint: To solve this question let the vertices of the tetrahedron are O$(0,0,0)$ , A$(a,0,0)$, B $(0,b,0)$and C $(0,0,c)$. The Volume of tetrahedron is V=\[\dfrac{1}{6}[\overrightarrow a \,\,\vec b\,\,\overrightarrow c ]\] .
Find the centroids of the each faces OAB, OAC, OBC,ABC are say ${H_1}\left( {\dfrac{a}{3},\dfrac{b}{3},0} \right)$ , ${H_2}\left( {\dfrac{a}{3},0,\dfrac{c}{3}} \right)$, ${H_3}$$\left( {0,\dfrac{b}{3},\dfrac{c}{3}} \right)$ and ${H_4}\left( {\dfrac{a}{3},\dfrac{b}{3},\dfrac{c}{3}} \right)$.
Since the distances are $\overrightarrow {{H_4}{H_1}} = \dfrac{{\overrightarrow c }}{3}$, $\overrightarrow {{H_4}{H_2}} = \dfrac{{\overrightarrow b }}{3}$,$\overrightarrow {{H_4}{H_3}} = \dfrac{{\overrightarrow a }}{3}$, Volume of tetrahedron by centroids,
$V' = \dfrac{1}{6}\left[ {\dfrac{{\overrightarrow a }}{3}\,\,\dfrac{{\overrightarrow b }}{3}\,\,\dfrac{{\overrightarrow c }}{3}} \right]$
By substituting V=\[\dfrac{1}{6}[\overrightarrow a \,\,\vec b\,\,\overrightarrow c ]\]into $V'$we can get the required relationship.

Complete step by step answer:
Consider vertices of the tetrahedron are O$(0,0,0)$ , A$(a,0,0)$, B $(0,b,0)$and C $(0,0,c)$.
 The Volume of tetrahedron is V=\[\dfrac{1}{6}[\overrightarrow a \,\,\vec b\,\,\overrightarrow c ]\] , where \[\overrightarrow a \,\], \[\vec b\], and \[\overrightarrow c \] are the distances $\overrightarrow {OA} $ , $\overrightarrow {OB} $,and $\overrightarrow {OC} $ respectively.
The centroid is given by coordinates: a triangle with vertices at \[\left( {{x_1},{\text{ }}{y_1},{z_1}} \right)\],\[\left( {{x_2},{\text{ }}{y_2},{z_2}} \right)\],\[\left( {{x_3},{\text{ }}{y_3},{z_3}} \right)\] has centroid at $\left( {\dfrac{{{x_1}{\text{ + }}{y_1} + {z_1}}}{3},\dfrac{{{x_2}{\text{ + }}{y_2} + {z_2}}}{3},\dfrac{{{x_3}{\text{ + }}{y_3} + {z_3}}}{3}} \right)$ .
Now, apply the centroids formula of faces of the tetrahedron O$(0,0,0)$ , A$(a,0,0)$, B $(0,b,0)$.
The centroids of the face OAB is found by substituting \[\left( {{x_1},{\text{ }}{y_1},{z_1}} \right) = (0,0,0)\],\[\left( {{x_2},{\text{ }}{y_2},{z_2}} \right) = (a,0,0)\]and \[\left( {{x_3},{\text{ }}{y_3},{z_3}} \right) = \left( {0,{\text{ }}b,0} \right)\].into the formula.
The centroid is ${H_1}\left( {\dfrac{a}{3},\dfrac{b}{3},0} \right)$.
 Similarly we can find centroids of faces OAC, OBC,ABC are , ${H_2}\left( {\dfrac{a}{3},0,\dfrac{c}{3}} \right)$, ${H_3}$$\left( {0,\dfrac{b}{3},\dfrac{c}{3}} \right)$ and ${H_4}\left( {\dfrac{a}{3},\dfrac{b}{3},\dfrac{c}{3}} \right)$respectively.
Since the distances are $\overrightarrow {{H_4}{H_1}} = \dfrac{{\overrightarrow c }}{3}$, $\overrightarrow {{H_4}{H_2}} = \dfrac{{\overrightarrow b }}{3}$,$\overrightarrow {{H_4}{H_3}} = \dfrac{{\overrightarrow a }}{3}$, Volume of tetrahedron by centroids,
$V' = \dfrac{1}{6}\left[ {\dfrac{{\overrightarrow a }}{3}\,\,\dfrac{{\overrightarrow b }}{3}\,\,\dfrac{{\overrightarrow c }}{3}} \right]$
By substituting V= \[\dfrac{1}{6}[\overrightarrow a \,\,\vec b\,\,\overrightarrow c ]\] into $V'$we can get the required relationship.
$V' = \dfrac{1}{6} \times \dfrac{1}{{27}}\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right]$
\[V' = \dfrac{1}{{27}}V\]
\[27V' = V\]
Here, the value of \[K\] is $27$ .

Note: Use Distance formula to find $\overrightarrow {{H_4}{H_1}} = \dfrac{{\overrightarrow c }}{3}$.
If ${H_1}\left( {\dfrac{a}{3},\dfrac{b}{3},0} \right)$and ${H_4}\left( {\dfrac{a}{3},\dfrac{b}{3},\dfrac{c}{3}} \right)$then distance,
$\overrightarrow {{H_4}{H_1}} = \sqrt {{{\left( {\dfrac{a}{3} - \dfrac{a}{3}} \right)}^2} + {{\left( {\dfrac{b}{3} - \dfrac{b}{3}} \right)}^2} + {{\left( {0 - \dfrac{c}{3}} \right)}^2}} $
$\overrightarrow {{H_4}{H_1}} = \sqrt {0 + 0 + {{\left( {\dfrac{c}{3}} \right)}^2}} $
$\overrightarrow {{H_4}{H_1}} = \dfrac{c}{3}$